#include<stdio.h>
int main()
{
    int x;
    double y;
    printf("请输入整数x;\n");
    scanf_s("%d", &x);
    y = (1 + x * x) / (x*x + 4 * sqrt(x) + 10);
    printf("y的值为：%lf\n", y);
    getchar();
    getchar();
}

//online parser: https://www.bccn.net/run/

#include <stdio.h>
#include <math.h>

int main(int argc, char *argv[])
{
    int x;
    double y;
    
    x = 10;
    y = (1 + x * x) / (x*x + 4 * sqrt(x) + 10);
    printf("y的值为：%lf\n", y);
    
    return 0;
}

output sample:
y的值为：0.823487

感谢感谢